Multivariable calculus

Limit of x^3y/(x^6+y^2): Parabolas Are Not Enough

Q: Why does the polar method fail to detect this?

In polar form f = \frac{r^2\cos^3\theta\sin\theta}{r^4\cos^6\theta + \sin^2\theta}, and for each fixed \theta with \sin\theta \neq 0 this tends to 0. Fixing \theta means approaching along a ray, so polar coordinates used that way are just the line test again.

Q: Is there a pattern connecting these textbook examples?

Yes. For \frac{x^n y}{x^{2n} + y^2} the killer path is always y = x^n, giving \tfrac{1}{2}, while every curve y = cx^k with k < n gives 0. Taking n = 2 gives the familiar \frac{x^2y}{x^4+y^2} and n = 3 gives this one.

The limit of x^3 y/(x^6 + y^2) at the origin does not exist. Every line through the origin gives 0 and so does every parabola y = ax^2, but the cubic curve y = x^3 gives 1/2 at every point. Matching the denominator's two terms is what finds the disagreeing path.

f(x,y) = \frac{x^3 y}{x^6 + y^2}

The limit does not exist

Path	Limit along it
along y = 0	0
along y = x	0
along the parabola y = x^2	0
along the cubic y = x^3	0.5

Escalating through the curves

Lines go first. Substituting $y = mx$ and cancelling $x^2$ from top and bottom leaves a factor that vanishes.

f(x, mx) = \frac{x^3 \cdot mx}{x^6 + m^2x^2} = \frac{mx^4}{x^2(x^4 + m^2)} = \frac{mx^2}{x^4 + m^2} \to 0

The vertical line $x = 0$ gives $0$ too, since its numerator is identically zero. Parabolas next: with $y = ax^2$ the same cancellation happens one power higher, and the result is still zero.

f(x, ax^2) = \frac{x^3 \cdot ax^2}{x^6 + a^2x^4} = \frac{ax^5}{x^4(x^2 + a^2)} = \frac{ax}{x^2 + a^2} \to 0

Now match the denominator. The two terms $x^6$ and $y^2$ are the same size when $y$ is comparable to $x^3$ , so try the cubic $y = x^3$ .

f(t, t^3) = \frac{t^3 \cdot t^3}{t^6 + (t^3)^2} = \frac{t^6}{2t^6} = \frac{1}{2}

The function is exactly $\tfrac{1}{2}$ everywhere on that curve. Lines give $0$ , the cubic gives $\tfrac{1}{2}$ , so the limit does not exist.

The systematic way to find the killer curve

Do not guess curves at random. Look at the denominator, set its two terms equal, and solve for one variable in terms of the other.

Denominator $x^6 + y^2$ : set $x^6 = y^2$ , so $y = \pm x^3$ .
Substitute that curve and see whether the numerator survives at the same order.
Here the numerator $x^3 y$ becomes $x^6$ , matching the denominator's $2x^6$ , so the ratio is a nonzero constant.
If the numerator came out at a higher order, the limit would be $0$ on that curve too and you would look for a proof instead.

The general family $y = cx^3$ gives $\frac{c}{1 + c^2}$ , which is $\tfrac{1}{2}$ at $c = 1$ , $-\tfrac{1}{2}$ at $c = -1$ , and tends to $0$ as $c$ grows. Any two different choices of $c$ already disprove the limit.

The mistake: stopping once lines and parabolas agree

Students who have learned that lines are not enough often upgrade to parabolas and stop there. This function defeats that too. There is no fixed family of curves whose agreement proves a limit exists, because for any degree $n$ you can build a function that only fails on degree $n+1$ curves.

The reliable rule: path tests can only ever disprove. Once several paths agree, switch to an argument that covers all approaches at once, such as bounding $|f|$ by something that goes to zero. Here no such bound exists, since $|f|$ reaches $\tfrac{1}{2}$ arbitrarily close to the origin.

The three-line habit that keeps you honest: substitute lines, look at the denominator for the balancing curve, then either produce a disagreement or produce an inequality. Never conclude from agreement alone.

Frequently asked questions

Why does the polar method fail to detect this?

In polar form $f = \frac{r^2\cos^3\theta\sin\theta}{r^4\cos^6\theta + \sin^2\theta}$ , and for each fixed $\theta$ with $\sin\theta \neq 0$ this tends to $0$ . Fixing $\theta$ means approaching along a ray, so polar coordinates used that way are just the line test again.

Is there a pattern connecting these textbook examples?

Yes. For $\frac{x^n y}{x^{2n} + y^2}$ the killer path is always $y = x^n$ , giving $\tfrac{1}{2}$ , while every curve $y = cx^k$ with $k < n$ gives $0$ . Taking $n = 2$ gives the familiar $\frac{x^2y}{x^4+y^2}$ and $n = 3$ gives this one.